repeated measure design of anova

Repeated Measure Design of ANOVA

AMS 572 Group 5

Outline

• Jia Chen: Introduction of repeated measures ANOVA

• Chewei Lu: One-way repeated measures • Wei Xi: Two-factor repeated measures• Tomoaki Sakamoto : Three-factor repeated

measures• How-Chung Liu: Mixed models• Margaret Brown: Comparison• Xiao Liu: Conclusion

Introduction of Repeated Measures ANOVA

Jia Chen

What is it ?

Definition: - It is a technique used to test the equality of

means.

When To Use It?

• It is used when all members of a random sample are measured under a number of different conditions.

• As the sample is exposed to each condition in turn, the measurement of dependent variable is repeated.

Introduction of One-Way Repeated Measures ANOVA

Che-Wei, LuProfessor:Wei Zhu

One-Way Repeated Measures ANOVA

• DefinitionA one-way repeated measures ANOVA instead of having one score per subject, experiments are frequently conducted in which multiple score are gathered for each case.

• Concept of Repeated Measures ANOVA One factor with at least two levels, levels are dependent. Dependent means that they share variability in some way. The Repeated Measures ANOVA is extended from standard

ANOVA.

• When to Use Measuring performance on the same variable over time– for example looking at changes in performance during training or

before and after a specific treatment The same subject is measured multiple times under

different conditions – for example performance when taking Drug A and performance

when taking Drug B The same subjects provide measures/ratings on

different characteristics– for example the desirability of red cars, green cars and blue cars

Note how we could do some RM as regular between subjects designs– For example, Randomly assign to drug A or B

One-Way Repeated Measures ANOVA• Source of Variance in Repeated Measures ANOVA SStotal

– Deviation of each individual score from the grand mean SSb/t subjects

– Deviation of subjects' individual means (across treatments) from the grand mean.

– In the RM setting, this is largely uninteresting, as we can pretty much assume that ‘subjects differ’

SSw/in subjects: How Ss vary about their own mean, breaks down into:– SStreatment • As in between subjects ANOVA, is the comparison of treatment means

to each other (by examining their deviations from the grand mean)• However this is now a partition of the within subjects variation

– SSerror• Variability of individuals’ scores about their treatment mean

• Partition of Sum of Square

𝑆𝑆 𝑡𝑜𝑡𝑎𝑙

𝑆𝑆𝑏 /𝑡𝑠𝑢𝑏𝑗𝑒𝑐𝑡𝑠 𝑆𝑆𝑤/ 𝑖𝑛𝑠𝑢𝑏𝑗𝑒𝑐𝑡𝑠

𝑆𝑆 𝑡𝑟𝑒𝑎𝑚𝑒𝑚𝑡 𝑆𝑆𝑒𝑟𝑟𝑜𝑟

Repeated Measures ANOVA

𝑆𝑆 𝑡𝑜𝑡𝑎𝑙

𝑆𝑆𝑏 /𝑡𝑡𝑟𝑒𝑎𝑚𝑒𝑛𝑡 𝑆𝑆𝑤 / 𝑖𝑛𝑒𝑟𝑟𝑜𝑟

Standard ANOVA

Variation SS Df MS FBetween a-1 MSA= F=

Within N-a MSE=

Total N-1

• Standard ANOVA Table

• Repeated Measures ANOVA TableSS Df MS F

Between a-1 MSA= F=

Within N-a

-Subjects s-1

-Error MSE=

Total N-A

One-Way Repeated Measures ANOVA• Example:Researchers want to test a new anti-anxiety medication. They measure the anxiety of 7 participants three times: once before taking the medication, once one week after taking the medication, and once two weeks after taking medication. Anxiety is rated on a scale of 1-10,with 10 being ”high anxiety” and 1 being “low anxiety”. Are there any difference between the three condition using significant level

Participants Before Week1 Week21 9 7 42 8 6 33 7 6 24 8 7 35 8 8 46 9 7 37 8 6 2

• Define Null and Alternative Hypotheses

• Define Degrees of Freedom N=21 s=7

-=18-6=12

• Analysis of Variance

Participants Before Week1 Week2

1 9 7 4

2 8 6 3

3 7 6 2

4 8 7 3

5 8 8 4

6 9 7 3

7 8 6 2

One-Way Repeated Measures ANOVA• Analysis of Variance(ANOVA Table)

SS Df MS F

Between 98.67 2 49.34 224.27

Within 10.29 18

-Subjects 7.62 6

-Error 2.67 12 0.22

Total 108.96 20

Error=within-Subjects=10.29-7.62=2.67Total=Beteen+Within=98.67+10.29=108.96

• Test Statistic:244.27

• Critical Region

Now, the

SS Df MS F

Between 98.67 2 49.34 224.27

Within 10.29 18

-Subjects 7.62 6

-Error 2.67 12 0.22

Total 108.96 20

One-Way Repeated Measures ANOVA• SAS Code DATA REPEAT;

INPUT SUBJ BEFORE WEEK1 WEEK2;DATALINES;1 9 7 42 8 6 33 7 6 24 8 7 35 8 8 46 9 7 37 8 6 2;PROC ANOVA DATA=REPEAT;TITLE "One-Way ANOVA using the repeated Statment";MODEL BEFORE WEEK1 WEEK2= / NOUNI;REPEATED TIME 3 (1 2 3);RUN;

The Before, Week1, and Week2 are the time level at each participants.

In here, we don’t have CLASS statement

because our data set does not have an

independent variable

The NOUNI(no univariate) is a request not to conduct a separate analysis for each of the three times variables.

This indicates the labels we want to printed for each level of times

One-Way Repeated Measures ANOVA• SAS Result

Two-Factor ANOVA with Repeated Measures

Wei Xi

Stating of the Hypothesis

Within-Subjects Main Effect

Between-Subjects Main Effect

Between-Subjects Interaction Effect

Within-Subjects By Between-Subjects Interaction Effects

Two-Factor ANOVA with Repeated Measures on One Factor

Hypothesis

ANOVA TABLESource DF SS MS F

Factor A a-1 SSA SSA/(a-1) MSA/MSWA ～ F(a-1),n(a-1)

Factor B b-1 SSB SSB/(b-1)MSB/MSE ～ F(b-1),n(a-1)(b-1)

AB Interaction (a-1)(b-1) SSAB SSAB/(a-1)(b-1)MSAB/MSE ～ F(a-1)(b-1,n(a-1)(b-1)

Subjects within A (n-1)a SSWA

SSWA/(n-1)a

Error (n-1)a(b-1) SSE SSE/((n-1)a(b-1)

Total nab-1 SST

Example• The shape variable is the repeated variable. This produces an ANOVA with one

between-subjects factor. If you were to examine the expected mean squares for this setup, you would find that the appropriate error term for the test of calib is subject|calib. The appropriate error term for shape and shape#calib is shape#subject|calib (which is the residual error since we do not include the term in the model).

SAS Code

Data Q1;

set pre.Q1;

proc anova data=Q1;

title' Two-way Anova with a Repeated Measure on One Factor';

class calib;

model shape_1 shape_2 shape_3 shape_4 = calib/nouni;

repeated shape 4;

means calib;

MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no shape EffectH = Anova SSCP Matrix for shape

E = Error SSCP Matrix

S=1 M=0.5 N=0Statistic Value F Value Num DF Den DF Pr > FWilks' Lambda 0.02529573 25.69 3 2 0.0377Pillai's Trace 0.97470427 25.69 3 2 0.0377Hotelling-Lawley Trace 38.53236607 25.69 3 2 0.0377Roy's Greatest Root 38.53236607 25.69 3 2 0.0377

MANOVA Test Criteria and Exact F Statistics for the Hypothesis of no shape*calib EffectH = Anova SSCP Matrix for shape*calib

E = Error SSCP Matrix

S=1 M=0.5 N=0Statistic Value F Value Num DF Den DF Pr > FWilks' Lambda 0.16750795 3.31 3 2 0.2404Pillai's Trace 0.83249205 3.31 3 2 0.2404Hotelling-Lawley Trace 4.96986607 3.31 3 2 0.2404Roy's Greatest Root 4.96986607 3.31 3 2 0.2404

At α=0.05,we reject the hypothesis and conclude that there is shape Effect

At α=0.05,we cannot reject the hypothesis and conclude that there is no shape*calib Effect

Analysis of SAS Output

Source DF Anova SS Mean Square F Value Pr > Fcalib 1 51.04166667 51.04166667 11.89 0.0261

Error 4 17.16666667 4.29166667

Source DF Anova SS Mean Square F Value Pr > F

Adj Pr > F

G - G H - Fshape 3 47.45833333 15.81944444 12.80 0.0005 0.0099 0.0011

shape*calib 3 7.45833333 2.48611111 2.01 0.1662 0.2152 0.1791

Error(shape) 12 14.83333333 1.23611111

Univariate Tests of Hypotheses for Within Subject Effects

Tests of Hypotheses for Between Subjects Effects

Two-Factor ANOVA with Repeated Measures on both Factors

Source DF SS MS F Subjects n-1 SSS SSS/I-1 MSS/MSEFactor A a-1 SSA SSA/(a-1) MSA/MSA*S ～ F(a-1),(n-1)(a-1)

Factor B b-1 SSB SSB/(b-1) MSB/MSB*S ～ F(b-1),(n-1)(b-1)

AB Interaction (a-1)(b-1) SSAB SSAB/((a-1)(b-1))

MSAB/MSE ～ F(a-1)(b-1),(n-1)(a-1)(b-1)

A*Subjects (n-1)(a-1) SSA*S SSWA/((n-1)a) SSA*S/MSEF(a-1)(n-1),(n-1)(a-1)(b-1)

B*Subjects (n-1)(b-1) SSB*S SSWB/((n-1)b) SSA*S/MSEF(n-1)(b-1),(n-1)(a-1)(b-1)

Error (n-a)(a-1)(b-1)

SSE SSE/((n-1)(a-1)(b-1))

Total nab-1 SST

ANOVA TABLE

Example Three subjects, each with nine accuracy scores on all combinations of the three

different dials and three different periods. With subject a random factor and both dial and period fixed factors, the appropriate error term for the test of dial is the dial#subject interaction. Likewise, period#subject is the correct error term for period, and period#dial#subject (which we will drop so that it becomes residual error) is the appropriate error term for period#dial.

SAS Code• Data Q2; • Input Mins1-Mins9; • Datalines; • 45 53 60 40 52 57 28 37 46 • 35 41 50 30 37 47 28 32 41 • 60 65 75 58 54 70 40 47 50 • ; • ODS RTF STYLE=BarrettsBlue; • Proc anova data=Q2; • Model Mins1-Mins9=/nouni; • Repeated period 3, dail 3/nom; • Run; • ods rtf close; •

SAS Output

Source DF Anova SS Mean Square F Value Pr > FAdj Pr > F

G - G H - Fperiod 2 1072.666667 536.333333 14.45 0.0148 0.0563 0.0394Error(period) 4 148.444444 37.111111

Greenhouse-Geisser Epsilon 0.5364

Huynh-Feldt Epsilon 0.6569

G - G H - Fdail 2 978.6666667 489.3333333 50.91 0.0014 0.0169 0.0115Error(dail) 4 38.4444444 9.6111111

G - G H - Fperiod*dail 4 8.66666667 2.16666667 0.30 0.8715 0.6603 0.7194

Error(period*dail) 8 58.22222222 7.27777778

Univariate Tests of Hypotheses for Within Subject Effects

At α=0.05,we cannot reject the hypothesis and conclude the there is no period*dail Effect

At α=0.05,we reject the hypothesis and conclude that there is dail Effect

At α=0.05,we reject the hypothesis and conclude that there is period Effect

Three-factor Experimentswith a repeated measure

T. Sakamoto

Example of a marketing experiment

Experiment The subjects who belong to a region X or Y see the

Liquid Crystal Display A, B, or C. Each type of LCD is seen twice; once in the light and

the other in the dark. The preferences of the LCD are measured by the

subjects, on a scale from 1 to 5 (1= lowest, 5=highest).

Case of this example• A company which produces some Liquid Crystal

Display wants to examine the characteristics of its prototype products.

Experimental Design and Data Three factors

Type of LCD Regions to which the specimens belong In the light / In the dark

Repeted measure factor : In the light / In the dark

Type of LCD

subj light dark subj light dark subj light dark

REGION

1 5 4 11 4 4 21 5 5

2 4 2 12 5 6 22 5 3

3 5 4 13 3 4 23 3 3

4 3 5 14 5 4 24 4 4

5 5 3 15 4 6 25 4 3

6 4 4 16 5 5 26 3 5

7 3 5 17 4 3 27 4 3

8 4 3 18 5 3 28 5 2

9 2 5 19 5 5 29 3 4

10 5 4 20 4 4 30 4 3

data lcd;input subj type $ region $ light dark @@;datalines;1 a 5 4 2 a 4 2 3 a 5 4 4 a 3 5 5 a 5 36 a 4 4 7 a 3 5 8 a 4 3 9 a 2 5 10 a 5 4 11 b 4 4 12 b 5 6 13 b 3 4 14 b 5 4 15 b 4 616 b 5 5 17 b 4 3 18 b 5 3 19 b 5 5 20 b 4 421 c 5 5 22 c 5 3 23 c 3 3 24 c 4 4 25 c 4 326 c 3 5 27 c 4 3 28 c 5 2 29 c 3 4 30 c 4 3;run;

proc anova data=lcd;title ’Three-way ANOVA with a Repeated Measure';class type region;model light dark = type | region /nouni;repeated light_dark;means type | region;run;

SAS PROGRAM

OUTPUT(Part 1/4):

OUTPUT(Part 2/4):

OUTPUT(Part 3/4):

OUTPUT(Part 4/4):

Mixed Effect ModelsHow-Chang Liu

Mixed Models

• When we have a model that contains random effect as well as fixed effect, then we are dealing with a mixed model.

• From the above definition, we see that mixed models must contain at least two factors. One having fixed effect and one having random effect.

Why use mixed models?

• When repeated measurements are made on the same statistical units, it would not be realistic to assume that these measurements are independent.

• We can take this dependence into account by specifying covariance structures using a mixed model

Definition

• A mixed model can be represented in matrix notation by:

• is the vector of observations• is the vector of fixed effects• is the vector of random effects• is the vector of I.I.D. error terms• and are matrices relating and to

Assumptions

• R and G are constants• We also assume that and are independent• We get V = ZGZ' + R, where V is the variance of

How to estimate and ?

If R and G are given: Using Henderson’s Mixed Model equation, we have: =

So = And = )

What if G and R are unknown?

• We know that both and are normally distributed, so the best approach is to use likelihood based methods

• There are two methods used by SAS:• 1)Maximum likelihood (ML)• 2)Restricted/residual maximum likelihood

(REML)

ExampleBelow is a table of growth measurements for 11 girls and 16 boys at ages 8, 10, 12, 14:

Person gender age8 age10 age12 age141 F 21.0 20.0 21.5 23.0 2 F 21.0 21.5 24.0 25.5 3 F 20.5 24.0 24.5 26.0 4 F 23.5 24.5 25.0 26.5 5 F 21.5 23.0 22.5 23.5 6 F 20.0 21.0 21.0 22.5 7 F 21.5 22.5 23.0 25.0 8 F 23.0 23.0 23.5 24.0 9 F 20.0 21.0 22.0 21.5 10 F 16.5 19.0 19.0 19.5 11 F 24.5 25.0 28.0 28.0 12 M 26.0 25.0 29.0 31.0 13 M 21.5 22.5 23.0 26.5 14 M 23.0 22.5 24.0 27.5

Person gender age8 age10 age12 age1415 M 25.5 27.5 26.5 27.016 M 20.0 23.5 22.5 26.0 17 M 24.5 25.5 27.0 28.5 18 M 22.0 22.0 24.5 26.5 19 M 24.0 21.5 24.5 25.5 20 M 23.0 20.5 31.0 26.0 21 M 27.5 28.0 31.0 31.5 22 M 23.0 23.0 23.5 25.0 23 M 21.5 23.5 24.0 28.0 24 M 17.0 24.5 26.0 29.5 25 M 22.5 25.5 25.5 26.0 26 M 23.0 24.5 26.0 30.0 27 M 22.0 21.5 23.5 25.0

Using SASdata pr; input Person Gender $ y1 y2 y3 y4; y=y1; Age=8; output; y=y2; Age=10; output; y=y3; Age=12; output; y=y4; Age=14; output; drop y1-y4; datalines; 1 F 21.0 20.0 21.5 23.0 2 F 21.0 21.5 24.0 25.5 …;Run;

Using SAS

proc mixed data=pr method=ml covtest; class Person Gender; model y = Gender Age Gender*Age / s; repeated / type=un subject=Person r; run;

ResultsModel Information

Data Set WORK.PR

Dependent Variable y

Covariance Structure Unstructured

Subject Effect Person

Estimation Method ML

Residual Variance Method None

Fixed Effects SE Method Model-Based

Degrees of Freedom Method Between-Within

Class Level Information

Class Levels Values

Person 27 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Gender 2 F M

As one can see, the covariance matrix is unstructured, as we are going to estimate it using the maximum likelihood method

Dimensions

Covariance Parameters 10

Columns in X 6

Columns in Z 0

Subjects 27

Max Obs Per Subject 4

Number of Observations

Number of Observations Read 108

Number of Observations Used 108

Number of Observations Not Used 0

Iteration History

Iteration Evaluations -2 Log Like Criterion

0 1 478.24175986

1 2 419.47721707 0.00000152

2 1 419.47704812 0.00000000

Convergence criteria met.

As one can see, we do not have a Z matrix for this model

the convergence of the Newton-Raphson algorithm means that we have found the maximum likelihood estimates

Covariance Parameter Estimates

Cov Parm Subject Estimate Standard Error Z Value Pr Z

UN(1,1) Person 5.1192 1.4169 3.61 0.0002

UN(2,1) Person 2.4409 0.9835 2.48 0.0131

UN(2,2) Person 3.9279 1.0824 3.63 0.0001

UN(3,1) Person 3.6105 1.2767 2.83 0.0047

UN(3,2) Person 2.7175 1.0740 2.53 0.0114

UN(3,3) Person 5.9798 1.6279 3.67 0.0001

UN(4,1) Person 2.5222 1.0649 2.37 0.0179

UN(4,2) Person 3.0624 1.0135 3.02 0.0025

UN(4,3) Person 3.8235 1.2508 3.06 0.0022

UN(4,4) Person 4.6180 1.2573 3.67 0.0001

The table lists the 10 estimated covariance parameters in order. In other words, these are the estimates for R, the variance of

Solution for Fixed Effects

Effect Gender Estimate Standard Error DF t Value Pr > |t|

Intercept 15.8423 0.9356 25 16.93 <.0001

Gender F 1.5831 1.4658 25 1.08 0.2904

Gender M 0 . . . .

Age 0.8268 0.07911 25 10.45 <.0001

Age*Gender F -0.3504 0.1239 25 -2.83 0.0091

Age*Gender M 0 . . . .

From this table, we see that the boys intercept is at 15.8423, whole the girls intercept is at 15.8423+1.5831=17.42.The estimate of the boys’ slope is at 0.827, while the girls’ slpe is at 0.827-0.350=0.477So the girls’ starting point is higher than the girls but their growth rate is only about half of that of the boys

Type 3 Tests of Fixed Effects

Effect Num DF Den DF F Value Pr > F

Gender 1 25 1.17 0.2904

Age 1 25 110.54 <.0001

Age*Gender 1 25 7.99 0.0091

This is probably the most important table from our results:

The gender row tests the null hypothesis that girls and boys have a common intercept.As we can see we cannot reject that hypothesis

The Age tests the null hypothesis that age does not affect the growth rate.As we can see, we reject the null hypothesis as the F-value is large.

The Age*gender tests reveals that there is a difference in slope at the 1% significance level.

Repeated Measures ANOVA vs. Independent Measures ANOVA

Magarate Brown

• Can we just use standard ANOVA with repeated measures data?

• No, Independent Measures (standard) ANOVA assumes the data are independent.

• Data from a repeated measures experiment not independent

How are standard ANOVA and repeated measures ANOVA the same?

• Independent measures ANOVA: an extension of the pooled variance t-test

• Repeated Measures ANOVA: an extension of the paired sample t-test

• Independent measures ANOVA: assumes the population variances are equal (homogeneity of variance)

• Repeated Measures ANOVA: sphericity assumption that the population variances of all the differences are equal

• Both assume Normality of the population

Advantages to using Repeated Measures instead of Independent Measures

• Limited number of subjects available• Prefer to limit the number of subjects• Less variability (finger tapping with caffeine

example)• Can examine effects over time

Drawbacks

• Practice effect• Example: subjects get better at performing a task

each time with “practice”

• Differential transfer: “This occurs when the effects of one condition persist and affect participants’ experiences during subsequent conditions.” (format: http://www.psychmet.com/id16.html)

• Example: medical treatments

Resources (to be formatted)

• http://www.psychmet.com/id16.html• http://www.utexas.edu/courses/schwab/sw38

8r7_spring_2007/SolvingProblemsInSPSS/Solving%20Repeated%20Measures%20ANOVA%20Problems.pdf

• http://en.wikipedia.org/wiki/Repeated_measures

• http://www.mhhe.com/socscience/psychology/shaugh/ch07_summary.html

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